Question

an investor deposited $6,160.00 in an investment account. the account pays 3.1% interest compounded annually. the investor leaves the money in the account for 2 years. the investor makes no additional deposits or withdrawals. what is the balance in the account at the end of the 2 years? $6,289.16 $6,478.26 $6,484.06 $6,686.21

Explanation:

Step1: Recall compound interest formula

The formula for compound interest is $A = P(1 + r)^t$, where $A$ is the amount of money accumulated after $n$ years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), and $t$ is the time the money is invested for in years.

Step2: Identify values

Given $P = 6100$, $r = 3.1\% = 0.031$, and $t = 2$.

Step3: Substitute into formula

Substitute the values into the formula: $A = 6100(1 + 0.031)^2$.
First, calculate $1 + 0.031 = 1.031$.
Then, calculate $(1.031)^2 = 1.031\times1.031 = 1.062961$.
Now, multiply by the principal: $A = 6100\times1.062961$.
Calculate $6100\times1.062961 = 6100\times1 + 6100\times0.062961 = 6100 + 384.0621 = 6484.0621\approx6484.06$.

Answer:

$\$6,484.06$ (corresponding to the option with this value)